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Old 12-30-2004, 11:23 PM   #1
CurlyA
What results from this breed?

I'm breeding a blizzard to a patternless. Does anyone know what will come of this? Thank you.
 
Old 12-31-2004, 12:04 AM   #2
StinaUIUC
normals that are double het for blizzard and patternless
 
Old 12-31-2004, 07:50 AM   #3
mmfrankford
Well since were asking this. What would the babies be of a pairing between a patternless and a Tremper albino?
 
Old 12-31-2004, 08:14 AM   #4
MatthewK
Any dominant x domninat trait will get you double hets.

Blizzard x Patternless = Dbl Het (Het Bilzzard, and Het Patternless)

Tremper x Patternless = Dbl Het (Het Tremper, and Het Patternless)

Dbl Het Tremper Patternless x Dbl Het Tremper Patternless = Chance for patternless albino trempers, though I dont remember the chances (x/out of x eggs)

I think that..

Dbl Het Blizzard Patternless x Dbl Het Blizzard Patternless = Chance for banana blizzards? Correct me if im wrong.

-Matthew
 
Old 12-31-2004, 09:21 AM   #5
lilroach56
Matthew you got it all right except for one thing

Dominant traits will show up in the F1 generation, recessive (albino, patternless, blizzard) show up in the F2 generation.

When you said "dominant" it should have been recessive.

Also the odds of getting a dbl homozygous from a dbl het x dbl het breeding are 1:15 (dbl homozygous:non dbl homozygous).

-note-
those odds are theoretical, over thousands amounts of trials the actual result will be close to that.
 
Old 12-31-2004, 10:28 AM   #6
StinaUIUC
The chances for a double recessive from a double het breeding is 1:16...not 1:15. Not a big difference, but a difference all the same On that note the chances of a double homozygous dominant (not showing either trait) is the same, the chance of being just albino or just patternless is 3:16 (each...of either is 6:16). 1:3 of each of the single recessive animals will be homozygous dominant for the other gene. The chance of being normal homozygous dominant at the patternless gene and het at the albino gene or homozygous dominant at the albino gene and het at the patternless gene would be 2:16 (each...of either is 4:16), and finally the chance of being double het is 4:16. This makes the total ratio of genotypes 1:2:2:4:1:2:1:2:1 (dbl hom dom:hom dom/het:het/hom dom:dbl het:hom rec/dom:hom rec/het: dom/hom rec:het/hom rec:dbl rec). Phenotypically the ratio is 9:3:3:1 (normal:rec 1:rec2:dbl rec).

I have no idea if that makes any sense whatsoever.....lol
 

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